Shooting hoops

Wow, Mark & Geoff's general solutions are brilliant. I never would have thought of that.

I solved it only considering the 83 made shots out of 100.

With 100 shots attempted, there would be 100 probabilities multiplied together. The probability for each made shot, as stated, is:

$P_{make} = \frac{makes + 1}{shots + 2}$

so the probability for each missed shot would be:

$P_{miss} = 1 - \frac{makes + 1}{shots + 2} = \frac{shots + 2 - (makes + 1)}{shots + 2} = \frac{(shots - makes) + 1}{shots + 2} = \frac{misses + 1}{shots + 2}$

where "makes," "misses," and "shots" represent the number of makes, misses, and shots taken prior to any given shot.

The number of prior shots taken will run from zero (at the first shot) to 99 (at the 100th shot). So the denominator of each probability (shots + 2, regardless of whether the shot is made or missed) will run from 2 to 101. Therefore, when all 100 probability fractions are multiplied, the denominator will be equal to 101!

At each of the 83 shots made, the number of prior shots made will run from zero (at the first shot made) to 82 (at the 83rd shot made). So 83 of the fractions will represent the probability that the shot is made, and those numerators (makes + 1) will run from 1 to 83. The locations in the sequence of those made shots will be dealt with later, but those numerators of 1 through 83 will be multiplied regardless of their locations, so when the 100 probability fractions are multiplied, there will be an 83! in the numerator.

Similarly, at each of the 17 shots missed, the number of prior shots missed will run from zero (at the first shot missed) to 16 (at the 17th shot missed). So 17 of the fractions will represent the probability that the shot is missed, and those numerators (misses + 1) will run from 1 to 17, which means the numerator will include 17! for the misses in addition to 83! for the makes.

So when all of the probabilities are multiplied, the result will be:

$\frac{83! 17!}{101!}$

Now we just have to account for all of the different places in the sequence the 17 misses and 83 makes can go. Out of 100 shots, we need to choose 17 slots for the misses, or 83 slots for the makes, both of which are equivalent:

$\binom{100}{17} = \binom{100}{83} = \frac{100!}{17! 83!}$

Then we multiply the product of the 100 probabilities by the combination of all possible locations of the makes and misses:

$\frac{83! 17!}{101!} * \frac{100!}{17! 83!} = \frac{83! 17! 100!}{101! 17! 83!} = \frac{1}{101}$

Now that I saw how each number of makes from 0 to 100 is equally probable, I could have arrived to that conclusion in the same way. The product of the probabilities would be:

$\frac{[(makes - 1) + 1]! [(misses - 1) + 1]!}{[(shots - 1) + 2]!} = \frac{makes! misses!}{(shots + 1)!}$

And the combination for the locations of the makes and misses:

$\binom{shots}{makes} or \binom{shots}{misses} = \frac{shots!}{makes! misses!}$

and the final result would be:

$\frac{makes! misses!}{(shots + 1)!} * \frac{shots!}{makes! misses!} = \frac{makes! misses! shots!}{(shots + 1)! makes! misses!} = \frac{1}{shots + 1}$

for any number of makes and total shots under the given conditions. So not nearly as elegant, and I'm sure I'm stating the obvious and using incorrect terminology, but that's how I did it so I figured I'd type it up.

If $P_{n,j}$ is the probability that Amanda has made $j$ shots after $n$ attempts, for $0 \le j \le n$ , then it is clear that $P_{1,0} = P_{1,1} = \tfrac12$ . Moreover, it is clear that $P_{n+1,0} \; = \; P_{n,0} \times \tfrac{n+1}{n+2} \qquad \qquad P_{n+1,n+1} \; = \; P_{n,n} \times \tfrac{n+1}{n+2}$ for all $n \ge 1$ , and hence it follows that $P_{n,0} \; = \; P_{n,n} \; = \; \tfrac{1}{n+1} \qquad \qquad n \ge 1 \;.$ More generally, if $1 \le j \le n$ then conditional probability arguments tell us that $P_{n+1,j} \; = \; P_{n,j} \times \tfrac{n-j+1}{n+2} + P_{n,j-1} \times \tfrac{j}{n+2}$ Arguing by induction, if we assume that $P_{n,j} = \frac{1}{n+1}$ for all $0 \le j \le n$ it follows that $P_{n+1,j} = \tfrac{1}{n+2}$ for all $1 \le j \le n$ , and hence for all $0 \le j \le n+1$ .

Thus we deduce that $P_{n,j} = \tfrac{1}{n+1}$ for all $0 \le j \le n$ . Since $P_{100,83} = \tfrac{1}{101}$ , the answer is $\boxed{102}$ .

Incidentally, this is another variant of the classic Polya's urn model. Two marbles, one black and one white, are in an urn. Every second, a marble is chosen from the urn at random, and returned together with another marble of the same colour. If $X_n$ is the number of white marbles in the urn after $n$ seconds, then $X_n$ has the same distribution as $B_n + 1$ , where $B_n$ is the number of hoops Amanda makes after $n$ attempts. With this insight, the fact that the distribution of $B_n$ is uniform is a standard result.

It turns out, that the added factor $P = \frac{b+1}{n+2}$ is enough to leave her with an equal probability of all possible outcomes. So, after 100 shots, each of the $101$ possible outcomes (making $0$ through $100$ shots) is equiprobable, so the probability of her making exactly $83$ shots is $1/101$ , and $1+101=\boxed{102}$

The probability of a single combinations is given by:

$\prod_{n=0}^{82}\frac{n+1}{n + 2} * \prod_{n=1}^{17}\frac{n}{n+2+82}$

There are $\binom{100}{83}$ ways to arrange the succesfull shots.

Hence the probability for all the combinations is:

$\prod_{n=0}^{82}\frac{n+1}{n + 2} * \prod_{n=1}^{17}\frac{n}{n+2+82}*\binom{100}{83}=\frac{1}{101}$

The answer is $1+101=\boxed{102}$

The solution is split into three steps :

1)Since the experiment will be conducted 100 times, no matter what the outcomes are, we can easily predict the denominator of the final term,

$D^{r} = (101)!$ , obtained by substituting n = 0, 1,.... , 100, in the denominator of the given expression.

2)A portion of numerator, like our lifes, comes from success, and some from failure. The successes will be dealt with first,

$N^r(i^{th}success) = b + 1$ ,

since we are looking at $i^{th}$ success, so far there must have been (i-1) successes. Hence, b = i-1, substituting,

$N^r(i^{th}success) = i$

So the 83 successes put togather, contribute,

$N_r(all successes) = 83!$

3) Now for the portion of the numerator contributed by failures,

$N^r(i^{th}failure) = N^r (1- \frac{b+1}{n+2})$ ,

$N^r(i^{th}failure) = n+2-(b+1)$

$N^r(i^{th}failure) = n - b+1$

On little deliberation, one can realise that (n-b) is the number of failures so far, also, we are trying to commit the $i^{th}$ failure, so there have been i-1 failures so far. Thus, n-b = i-1.

$N^r(i^{th}failure) = i$

Since we have 17 failures in total,

$N^r(all failures) = 17!$

The probability that 17 failures and 83 successes occur in a particular order is,

$P(83-S, 17-F) = \frac{83!.17!}{101!}$

The total number of arrangements that can lead to this case is 100 choose 17. Thus the required prob is,

$P = \frac{100!}{83!.17!}.\frac{83!.17!}{101!} = \frac{1}{101}$